Asked  7 Months ago    Answers:  5   Viewed   33 times

please check the following code.

$imagebaseurl = 'support/content_editor/uploads/$name';

The $imagebaseurl is a variable that is containing a link to my image folder (uploads) and inside the folder I have some other folders which are named after my users name. for example: I have a user who's name is john, so the the link should look like this-> support/content_editor/uploads/john.

The main idea is when any user is logged in and browses his image gallery I want to take him to his own gallery which basically is named after his name.

When he will visit the gallery the value of $name in the link will come from the user's login name (from session). Now the problem is as you probably have already understood that the placement of $name in the above link is wrong and that is why it is not working. I am getting this whole URL> (support/content_editor/uploads/$name) instead of (support/content_editor/uploads/john)

Now could you please tell me how to use the $name in this $imagebaseurl = 'support/content_editor/uploads/$name';


$imagebaseurl = 'support/content_editor/uploads/' . $name;


$imagebaseurl = "support/content_editor/uploads/{$name}";

Note that if you use double quotes, you can also write the above as:

$imagebaseurl = "support/content_editor/uploads/$name";

It's good though to get in the habit of using {$...} in double quotes instead of only $..., for times where you need to insert the variable in a string where it's not obvious to PHP which part is the variable and which part is the string.

If you want the best performance, use string concatenation with single quotes.

Wednesday, March 31, 2021
answered 7 Months ago

You should point to your vendor/autoload.php at Settings | PHP | PHPUnit when using PHPUnit via Composer.

This blog post has all the details (with pictures) to successfully configure IDE for such scenario:

Related usability ticket:

P.S. The WI-18388 ticket is already fixed in v8.0

Wednesday, March 31, 2021
answered 7 Months ago

On Mac OS X environment variables available in Terminal and for the normal applications can be different, check the related question for the solution how to make them similar.

Note that this solution will not work on Mountain Lion (10.8).

Saturday, May 29, 2021
answered 5 Months ago
$query = "INSERT INTO `" . $mysql_tb . "`
VALUES ('".$studno."','".$firstname."','".$lastname."')";
Monday, August 16, 2021
answered 2 Months ago

If you are working with PHP 5.2, you can use call_user_func (or call_user_func_array) :

$className = 'A';

call_user_func(array($className, 'method'));

class A {
    public static function method() {
        echo 'Hello, A';

Will get you :

Hello, A

The kind of syntax you were using in your question is only possible with PHP >= 5.3 ; see the manual page of Static Keyword, about that :

As of PHP 5.3.0, it's possible to reference the class using a variable. The variable's value can not be a keyword (e.g. self, parent and static).

Wednesday, August 18, 2021
answered 2 Months ago
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